Example

Choosing the Appropriate Special Product Pattern for (9b−2)(2b+9)(9b - 2)(2b + 9), (9p−4)2(9p - 4)^2, (7y+1)2(7y + 1)^2, and (4r−3)(4r+3)(4r - 3)(4r + 3)

Choose the correct pattern—Binomial Squares, Product of Conjugates, or general FOIL—for each product, then compute the result.

ⓐ (9b−2)(2b+9)(9b - 2)(2b + 9): The two binomials do not share the same pair of first and last terms, so neither special product pattern applies. Use the FOIL method: (9b−2)(2b+9)=18b2+81b−4b−18=18b2+77b−18(9b - 2)(2b + 9) = 18b^2 + 81b - 4b - 18 = 18b^2 + 77b - 18

ⓑ (9p−4)2(9p - 4)^2: A single binomial is squared, so use the Binomial Squares Pattern (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, with a=9pa = 9p and b=4b = 4: (9p−4)2=(9p)2−2(9p)(4)+42=81p2−72p+16(9p - 4)^2 = (9p)^2 - 2(9p)(4) + 4^2 = 81p^2 - 72p + 16

ⓒ (7y+1)2(7y + 1)^2: Again, a single binomial is squared. Use (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, with a=7ya = 7y and b=1b = 1: (7y+1)2=(7y)2+2(7y)(1)+12=49y2+14y+1(7y + 1)^2 = (7y)^2 + 2(7y)(1) + 1^2 = 49y^2 + 14y + 1

ⓓ (4r−3)(4r+3)(4r - 3)(4r + 3): The binomials share the same first term 4r4r and the same last term 33, with one using subtraction and the other addition; they are conjugates. Apply the Product of Conjugates Pattern (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2, with a=4ra = 4r and b=3b = 3: (4r−3)(4r+3)=(4r)2−32=16r2−9(4r - 3)(4r + 3) = (4r)^2 - 3^2 = 16r^2 - 9

0

1

Updated 2026-06-03

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.5 Polynomials and Polynomial Functions - Intermediate Algebra @ OpenStax

Algebra

Related